# Tag Archives: italiano

## is it enough…

that the presentation given by

$\langle A,B\ :\ A^2=B^3,\quad A^2B=BA^2,\quad A^4=e,\quad B^6=e\rangle$

determines the group $SL_2({\mathbb{Z}}/{3\mathbb{Z}})$?

Similar question for

$\langle A,B,C$:

$A^2=B^3$,

$A^2B=BA^2$,

$A^4=B^6= C^2=e$,

$AC=CA,BC=CB\rangle$

for the group $GL_2({\mathbb{Z}}/{3\mathbb{Z}})$.

Other similar problems but less “difficult” are:

• $\langle\varnothing:\varnothing\rangle=\{e\}$
• $\langle A\ :\ A^2=e\rangle$  for  $\mathbb{Z}_2$
• $\langle A\ :\ A^3=e\rangle$  for  $\mathbb{Z}_3$
• $\langle A\ :\ A^4=e\rangle$  for  $\mathbb{Z}_4$
• $\langle A,B\ :\ A^2=e, B^2=e, AB=BA\rangle$  for  $\mathbb{Z}_2\oplus\mathbb{Z}_2$
• $\langle A,B\ :\ A^2=e, B^3=e, AB=B^2A\rangle$  for  $S_3$
• $\langle A,B\ :\ A^2=e, B^3=e, AB=BA\rangle$  for  $\mathbb{Z}_2\oplus\mathbb{Z}_3$
• $\langle A:\varnothing\rangle$ it is $\mathbb{Z}$
• $\langle A,B:AB=BA\rangle$  is $\mathbb{Z}\oplus\mathbb{Z}$
• $\langle A,B:\varnothing\rangle$  is $\mathbb{Z}*\mathbb{Z}$, the rank two free group
• $\langle A,B,J:(ABA)^4=e,ABA=BAB\rangle$ for $SL_2(\mathbb{Z})$
• $\langle A,B,J:(ABA)^4=J^2=e,ABA=BAB, JAJA=JBJB=e \rangle$ for $GL_2(\mathbb{Z})$
• $\langle A,B:A^2=B^3=e \rangle$ for $P\!S\!L_2(\mathbb{Z})=SL_2(\mathbb{Z})/{\mathbb{Z}_2}\cong{\mathbb{Z}}_2*{\mathbb{Z}}_3$
• dare altri venti esempi
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Filed under algebra, free group, group theory, mathematics